Among Be, B, Mg and Al, the second ionisation | vedclass

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(A) The electronic configurations of the given elements are:$Be (Z=4): 1s^2 2s^2$$B (Z=5): 1s^2 2s^2 2p^1$$Mg (Z=12): [Ne] 3s^2$$Al (Z=13): [Ne] 3s^2

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$B$

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(A) The electronic configurations of the given elements are:$Be (Z=4): 1s^2 2s^2$$B (Z=5): 1s^2 2s^2 2p^1$$Mg (Z=12): [Ne] 3s^2$$Al (Z=13): [Ne] 3s^2 3p^1$To find the second ionisation potential, we look at the configuration of the $+1$ ions:$Be^+: 1s^2 2s^1$$B^+: 1s^2 2s^2$$Mg^+: [Ne] 3s^1$$Al^+: [Ne] 3s^2$Removing the second electron from $B^+$ $(2s^2)$ is the most difficult because it involves removing an electron from a completely filled $2s$ subshell, which is highly stable. Therefore, $B$ has the maximum second ionisation potential.

Among $Be$, $B$, $Mg$ and $Al$, the second ionisation potential is maximum for

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